Merge pull request #432 from Krishna200608/feature/issue-378-nearest-smaller-values

Solved Issue #378: Added solution for Nearest Smaller Values

Author: Dheeraj-06

Problems/Data-Structures/Day-05/sol/Krishna200608/Solution1.cpp

@@ -0,0 +1,76 @@
+/*
+Problem: Nearest Smaller Values
+Link: https://cses.fi/problemset/task/1645
+Author: Krishna Sikheriya (Krishna200608)
+
+Short Problem Statement:
+Given an array of n integers, find for each array position the nearest position to its left having a smaller value. If no such position exists, print 0.
+
+Approach:
+We use a Monotonic Stack to solve this in linear time. 
+The stack will store pairs of (value, index) and maintain an increasing order of values.
+For each element in the array:
+1. We pop elements from the stack that are greater than or equal to the current element, as they cannot be the nearest smaller value for the current or subsequent elements.
+2. If the stack becomes empty, it means there is no smaller value to the left; we print 0.
+3. If the stack is not empty, the top element is the nearest smaller value; we print its index.
+4. We push the current element and its index onto the stack.
+
+Time Complexity: O(n) - Each element is pushed and popped from the stack at most once.
+Space Complexity: O(n) - In the worst case (increasing array), the stack stores all elements.
+
+Example I/O:
+Input:
+8
+2 5 1 4 8 3 2 5
+
+Output:
+0 1 0 3 4 3 3 7
+*/
+
+#include <iostream>
+#include <vector>
+#include <stack>
+
+using namespace std;
+
+int main() {
+    // Optimization for faster I/O operations
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    int n;
+    if (!(cin >> n)) return 0;
+
+    vector<int> a(n);
+    for (int i = 0; i < n; ++i) {
+        cin >> a[i];
+    }
+
+    // Stack stores pairs of {value, 1-based index}
+    stack<pair<int, int>> s; 
+    
+    for (int i = 0; i < n; ++i) {
+        // Pop elements greater than or equal to current element
+        // We only care about smaller values to the left
+        while (!s.empty() && s.top().first >= a[i]) {
+            s.pop();
+        }
+
+        if (s.empty()) {
+            cout << 0 << " ";
+        } else {
+            cout << s.top().second << " ";
+        }
+
+        // Push current element (value, index) to stack
+        s.push({a[i], i + 1});
+    }
+    cout << endl;
+
+    return 0;
+}
+
+/*
+SUBMISSION LINK: 
+https://cses.fi/paste/b62e11edd0054b5cf11189/
+*/